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Question:

A student measures the distance traversed in free fall of a body, initially at rest, in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e1 and e2 respectively, the percentage error in the estimation of g is:

e2−e1

e1+2e2

e1+e2

e1𕒶e2

Solution:

From the relation: h=ut+1/2gt²
h=1/2gt² (∵body initially at rest)
∴g=2h/t²
Taking logarithm on both sides,
log g = log 2 + log h - 2log t
Differentiating,
Δg/g = Δh/h - 2Δt/t
Percentage error in g = Percentage error in h - 2(Percentage error in t)
= e1 - 2e2
Percentage error in g = e1 + 2e2